1 / 7

Supplement Partial Derivatives

Supplement Partial Derivatives. 1-D Derivatives (Review). Given f ( x ) . f' ( x ) = d f /d x = Rate of change of f at point x w.r.t. x. f ( x + h ) – f ( x ). h. x. x+h. 1 st -order Partial Derivatives. For a function with 2 variables, f ( x , y ).

anne-hays
Télécharger la présentation

Supplement Partial Derivatives

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. SupplementPartial Derivatives

  2. 1-D Derivatives (Review) Given f(x). f'(x) = df/dx = Rate of change of f at point x w.r.t. x f(x + h) – f(x) h x x+h

  3. 1st-order Partial Derivatives For a function with 2 variables, f(x, y)

  4. Evaluating Partial Derivatives To find fx, we differentiate f w.r.t. x and treat all other variables as constants. Example 1: f(x, y) = x3y2 = (y2)x3 To evaluate fx, y2is to be treated as a constant term. Thus fx = (y2)(3x2) = 3x2y2 Example 2: f(x, y, z) = 10xyz + sin(y)exz3 = (10yz)x + (sin(y)z3)ex To evaluate fx, (10yz) and (sin(y)z3) are to be treated as constant terms. Thus fx = 10yz + (sin(y)z3)ex

  5. Exercise 1) f(x, y) = xexy • fx = • fy = 2) f(x, y, z) = sin (xy) • fx = • fy = • fz =

  6. 2nd-order Partial Derivatives

  7. Exercise f(x, y) = x3 + x2 y – 3xy2 + y3 • fx = • fy = • fxy = • fyx =

More Related